Proof: We go by induction on n. Consider the pair (Sn,A) where A is the equatorial sphere.
f defines a map

f~:ℝ⁢Pn→ℝ⁢Pn

. By cellular approximation, this may be
assumed to take the hyperplane at infinity (the n-1-cell of the standard cell structure on
ℝ⁢Pn) to itself. Since whether a map lifts to a covering depends only on its homotopy
class, f is homotopic to an odd map taking A to itself. We may assume that f is such a map.

is
not trivial. I claim it is an isomorphism. Hn⁢(Sn,A;ℤ2) is generated by cycles [R+] and
[R-] which are the fundamental classes of the upper and lower hemispheres, and the antipodal
map exchanges these. Both of these map to the fundamental class of A,
[A]∈Hn-1⁢(A;ℤ2). By the commutativity of the diagram,
∂⁡(f*⁢([R±]))=f*⁢(∂⁡([R±]))=f*⁢([A])=[A]. Thus f*⁢([R+])=[R±] and f*⁢([R-])=[R∓] since f commutes with the antipodal map. Thus f* is an isomorphism on
Hn⁢(Sn,A;ℤ2). Since Hn⁢(A,ℤ2)=0, by the exactness of the sequencei:Hn⁢(Sn;ℤ2)→Hn⁢(Sn,A;ℤ2) is injective, and so by the commutativity of the diagram (or equivalently
by the 5-lemma) f*:Hn⁢(Sn;ℤ2)→Hn⁢(Sn;ℤ2) is an isomorphism. Thus
f has odd degree.

The other statement of the Borsuk-Ulam theorem is:

There is no odd map Sn→Sn-1.

Proof: If f where such a map, consider f restricted to the equator A of Sn. This is an odd
map from Sn-1 to Sn-1 and thus has odd degree. But the map

f*⁢Hn-1⁢(A)→Hn-1⁢(Sn-1)

factors through Hn-1⁢(Sn)=0, and so must be zero. Thus f|A has degree 0, a
contradiction.